from what I saw, integration by parts was the only option. so i went ahead and let u = x and dv/dx = e^(x^2 /2). So i ended up getting du/dx = 1 and v = (e^(x^2 /2))/2 <--- however im not too sure if i got v correct since my CAS does not give me any answer when I integrate dv/dx.
Experts, please help
Experts, please help
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∫ x e^(x²/2) dx
Put x² / 2 = u , => x dx = du
The given integral reduces to ∫ e^u du = e^u + const
= e^(x²/2) + const [ Restoring the value of u ]
Put x² / 2 = u , => x dx = du
The given integral reduces to ∫ e^u du = e^u + const
= e^(x²/2) + const [ Restoring the value of u ]
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Try u-substitution with x^2/2 equaling u and du=x
integrate e^u du
integrate e^u du